Question: Let $g(x)=\begin{cases} (x-2)^2&\text{for }x \leq 2 \\\\ 2-x^2&\text{for }x>2 \end{cases}$ Is $g$ continuous at $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: For $g$ to be continuous at $x=2$, we need $\lim_{x\to 2}g(x)$ and $g(2)$ to exist and be equal. Since $2\leq 2$, the rule that applies to $x=2$ is $(x-2)^2$. So $g(2)=(2-2)^2=0$. Now let's analyze $\lim_{x\to 2}g(x)$. Finding $\lim_{x\to 2^{ +}}g(x)$ For $x$ -values larger than $2$, the appropriate rule for $g(x)$ is $2-x^2$. Since $2-x^2$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 2^{ +}}g(x) \\\\ &=\lim_{x\to 2^{ +}}[2-x^2] \gray{2-x^2\text{ is the rule for }x>2} \\\\ &=2-2^2 \gray{2-x^2\text{ is continuous at }x=2} \\\\ &=-2 \end{aligned}$ Finding $\lim_{x\to 2^{ -}}g(x)$ For $x$ -values smaller than $2$, the appropriate rule for $g(x)$ is $(x-2)^2$. Since $(x-2)^2$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 2^{ -}}g(x) \\\\ &=\lim_{x\to 2^{ -}}[(x-2)^2] \gray{(x-2)^2\text{ is the rule for }x<2} \\\\ &=(2-2)^2 \gray{(x-2)^2\text{ is continuous at }x=2} \\\\ &=0 \end{aligned}$ Conclusion We found that $\lim_{x\to 2^{ +}}g(x)=-2$ and $\lim_{x\to 2^{ -}}g(x)=0$. Since the one-sided limits aren't equal, $\lim_{x\to 2}g(x)$ doesn't exist and $g$ isn't continuous at $x=2$.